What is the integral of $x e^{x}$ $xe^x$ from -infinity to 0?

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What is the integral of $x e^{x}$ $xe^x$ from -infinity to 0?

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Answer 1 · 2015-07-14T16:24:33

I found: $\int_{- \infty}^{0} x e^{x} d x = - 1$ $int_(-oo)^0xe^xdx=-1$

Explanation:

I would solve it by parts:
$\int x e^{x} d x = x e^{x} - \int 1 \cdot e^{x} d x = x e^{x} - e^{x} =$ $intxe^xdx=xe^x-int1*e^xdx=xe^x-e^x=$
using your extrema:
$= e^{x} (x - 1) {∣ ∣}_{- \infty}^{0}$ $=e^x(x-1)|_-oo^0$ $=$ $=$
$= - 1 - 0 = - 1$ $=-1-0=-1$

[where I used the fact that: $e^{0} = 1 and e^{- \infty} = \frac{1}{e^{\infty}} = 0$ $e^0=1 and e^-oo=1/e^oo=0$ ]

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What is the integral of $x e^{x}$ $xe^x$ from -infinity to 0?

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What is the integral of $x e^{x}$ $xe^x$ from -infinity to 0?